How do you find the derivative of s=(2t+1)(3t-2)?

Jun 1, 2017

Answer: $s ' = 12 t - 1$

Explanation:

Differentiate $s = \left(2 t + 1\right) \left(3 t - 2\right)$

Consider the product rule:
$h \left(x\right) = f \left(x\right) \cdot g \left(x\right)$
$h ' \left(x\right) = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) g ' \left(x\right)$

So, we have that:
$s ' = \left(2\right) \left(3 t - 2\right) + \left(2 t + 1\right) \left(3\right)$
$s ' = 6 t - 4 + 6 t + 3$
$s ' = 12 t - 1$

Jun 2, 2017

$\frac{\mathrm{ds}}{\mathrm{dt}} = 12 t - 1$

Explanation:

The other answer is completely valid, I just thought I would show the other method of differentiation in this case.

First, use the FOIL method (from all the way back in Algebra 1) to expand the polynomial.

$s = \left(2 t + 1\right) \left(3 t - 2\right)$

$s = 6 {t}^{2} - 4 t + 3 t - 2$

$s = 6 {t}^{2} - t - 2$

Now, differentiate using power rule. No product rule required!

$\frac{\mathrm{ds}}{\mathrm{dt}} = 12 t - 1$